Question 210724
Using the Pythagorean theorem: {{{c^2 = a^2+b^2}}} where c (the hypotenuse) = 26, we can write:
1) {{{26^2 = a^2+b^2}}} and...
2) {{{a+b = 34}}} "The sum of the legs...is 34 meters." Rewrite this as:
{{{a = 34-b}}} and substitute into equation 1).
1a) {{{26^2 = (34-b)^2+b^2}}} Simplify and solve for b.
1b) {{{676 = (1156-68b+b^2)+b^2}}}
1c) {{{676 = 1156-68b+2b^2}}} Rearrange into the standard form for a quadratic equation.
1d) {{{2b^2-68b+480 = 0}}} Divide by 2 to ease the calculation a bit.
1e) {{{b^2-34b+240 = 0}}} Factor this trinomial.
1f) {{{(b-24)(b-10) = 0}}}  so...
{{{b = 24}}} or {{{b = 10}}}
So the two legs are:
a = 24 and b = 10 (or vice versa if you like)
Check:
{{{26^2 = 24^2+10^2}}}
{{{676 = 576+100}}}
{{{676 = 676}}}